Yesterday ago we learnt about surface integrals, and I already calculated some with parameters.
One of these exercises was this one:
$x= v\cos(u)$
$y= \sin(u)$
$z= v$
while $B' = {(u,v): 0 \le u \le \pi, 0 \le v \le h\sin(u))}$
this was very simple, as we were given a relatively solid explanation of how to calculate the surface distortion factor N by calculating $\sqrt{A²+B²+C²}$, while A,B,C are the corresponding Iacobi matrices. (I hope this is an international known way to calculate this - I am studying in Germany).
However, now I am facing an exercise where all I am given is:
$z = x*y$ as an area somewhere in my coordinate system.
Now, I have to calculate the surface of this area, but only of the part which is located above $x²+y² \le 1$.
I am pretty sure what I am calculating, but I just can't find a way to push this into a format which is applicable for the formulas I know.
Currently, I am trying to do it like this:
$u = x$
$v = y$
However, I think this is somewhat too idiotic, and that there is a special way to do it without parameters.
Can anyone of you help me? My main problem right now is to calculate the surface distortion factor. I think calculating the integral itself later on is pretty straightforward, considering the use of polar coordinates which come to my mind when reading about the given circle $x² + y² \le 1$